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Answer = D. \(75+25\sqrt{3}\)

Refer diagram below:

Attachment:
gpp-sgf_img3.png
gpp-sgf_img3.png [ 9.49 KiB | Viewed 4511 times ]

Sides of right triangle 30:60:90 are in the ratio \(1: \sqrt{3}:2\)

So the dimensions of the sides would be as shown in the diagram

Area of trapezoid \(= \frac{1}{2} (5 + 5 + 5\sqrt{3} + 5\sqrt{3}) * 5\sqrt{3} = 75+25\sqrt{3}\)
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Bunuel

In the diagram, HJLM is a square, and GH = 10. Find the area of trapezoid GHJK.

A. \(50+25\sqrt{3}\)

B. \(50+50\sqrt{3}\)

C. \(75+12.5\sqrt{3}\)

D. \(75+25\sqrt{3}\)

E. \(75+50\sqrt{3}\)

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MAGOOSH OFFICIAL SOLUTION:

This entire problem hinges on recognizing that triangles GHM and JKL are 30-60-90 triangles and using the properties of those triangles. First of all, notice that because HJLM is a square, it must be true that HM = JL. This means, the two triangles, GHM and JKL, must be congruent and have all the same sides & angles. GH = JK = 10.

Now, in a 30-60-90 triangle, the smaller leg is half the hypotenuse, so GM = LK = 5. The longer leg is the square root of 3 times the shorter leg, so

\(HM = JL = 5\sqrt{3}\)

This is the side of the square, so we square this to get the area of square HJLM.

\(Area \ of \ HJLM = (5\sqrt{3})^2=75\)

That’s part of the area. Now, we need the area of the triangles. One triangle is A = 0.5bh, so two triangles would simply be A = bh

\(Area \ of \ triangle=bh =GM*HM=5*5\sqrt{3}=25\sqrt{3}\)

Total area = square + triangles = \(75 + 25\sqrt{3}\).

Answer = (D)
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